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प्रश्न
Prove the following trigonometric identities.
`(1 + sin θ)/cos θ+ cos θ/(1 + sin θ) = 2 sec θ`
Prove the following:
`(1 + sin θ)/cos θ+ cos θ/(1 + sin θ) = 2 sec θ`
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उत्तर १
We have to prove `(1 + sin θ)/cos θ + cos θ/(1 + sin θ) = 2 sec θ`
We know that, `sin^2 θ + cos^2 θ = 1`
Multiplying the denominator and numerator of the second term by (1 − sin θ), we have
= `(1 + sin θ)/cos θ + cos θ/(1 + sin θ)`
`(1 + sin θ)/cos θ = (cos θ(1 - sin θ))/((1 + sin θ)(1 - sin θ))`
`(1 + sin θ)/cos θ = (cos θ (1 - sin θ))/(1-sin θ)`
= `(1 + sin θ)/cos θ + (cos θ(1 - sin θ))/cos^2 θ`
= `(1 + sin θ)/cos θ + (1 - sin θ)/cos θ`
= `(1 + sin θ + 1 - sin θ)/cos θ`
`= 2/cos θ`
= 2 sec θ
उत्तर २
LHS = `(1 + sin θ)/cos θ + cos θ/(1 + sin θ)`
= `(( 1 + sin θ)^2 + cos^2 θ)/(cos θ(1 + sin θ))`
= `(1 + sin^2 θ + 2 sin θ + cos^2 θ)/(cos θ(1 + sin θ ))`
= `(1 + (sin^2θ + cos^2 θ) + 2 sin θ)/(cos θ(1 + sin θ))`
= `(1 + 1 + 2sin θ)/(cos θ(1 + sin θ))`
= `(2(1 + sin θ))/(cos θ(1 + sin θ))`
= 2 sec θ
Hence proved.
संबंधित प्रश्न
If `sin theta = x , " write the value of cot "theta .`
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
What is the value of \[\frac{\tan^2 \theta - \sec^2 \theta}{\cot^2 \theta - {cosec}^2 \theta}\]
Write True' or False' and justify your answer the following:
\[ \cos \theta = \frac{a^2 + b^2}{2ab}\]where a and b are two distinct numbers such that ab > 0.
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Choose the correct alternative:
1 + tan2 θ = ?
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
If A + B = 90°, show that `(sin B + cos A)/sin A = 2tan B + tan A.`
Prove the following identities: cot θ - tan θ = `(2 cos^2 θ - 1)/(sin θ cos θ)`.
If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.
